Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 100 + 15;
const int inf = 0x3f3f3f3f;
int n, m, mat[maxn][maxn], ans = inf, coladd[maxn];
vector<pair<int, int> > Final;
int main(int argc, char* argv[]) {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) mat[i][j] = read();
  int mi = inf;
  for (int i = 1; i <= m; ++i) mi = min(mi, mat[1][i]);
  for (int wr = 0; wr <= mi; ++wr) {
    int sum = 0;
    vector<pair<int, int> > newopr;
    for (int i = 1; i <= wr; ++i) newopr.emplace_back(make_pair(0, 1));
    for (int i = 1; i <= m; ++i) {
      coladd[i] = mat[1][i] - wr;
      sum += coladd[i];
    }
    int ok = 1;
    for (int i = 2; i <= n && ok; ++i) {
      set<int> fw;
      for (int j = 1; j <= m && fw.size() <= 1 && ok; ++j) {
        if (coladd[j] > mat[i][j]) {
          ok = 0;
          break;
        }
        fw.insert(mat[i][j] - coladd[j]);
      }
      if (fw.size() > 1) {
        ok = 0;
        break;
      } else {
        int ww = *fw.begin();
        for (int k = 1; k <= ww; ++k) newopr.emplace_back(make_pair(0, i));
      }
    }
    if (!ok) continue;
    if (ok && sum + newopr.size() < ans) {
      ans = sum + newopr.size();
      for (int i = 1; i <= m; ++i)
        for (int j = 1; j <= coladd[i]; ++j)
          newopr.emplace_back(make_pair(1, i));
      Final = newopr;
    }
  }
  if (ans == inf)
    puts("-1");
  else {
    printf("%d\n", ans);
    for (auto it : Final) {
      if (it.first == 0)
        printf("row %d\n", it.second);
      else
        printf("col %d\n", it.second);
    }
  }
  return 0;
}